# Homework Help: Equations for instantaneous displacement of a particle

1. Oct 7, 2015

### AJDangles

1. The problem statement, all variables and given/known data
For each expression of displacement (x) below, write down the mathematical functions for velocity (v), acceleration (a), and pressure (Pa). Assume a maximum amplitude (A) of displacement = 1. Units are not required. Hint: Think about the phase relationships of pendular motion. You can express your answers by using a different trigonometric function or by simply changing the phase of the given function.

If the instantaneous displacement of particle x can be described by the function x = sin(ωt + θ), then write down the mathematical functions for v, a, and Pa.

v =

a =

Pa =

3. The attempt at a solution
I really don't have much of an idea of where to begin. I think it has something to do with a time domain waveform, but a nudge in the right direction would really help!

2. Oct 7, 2015

### haruspex

What, in complete generality, is the relationship between displacement, velocity and time?

3. Oct 7, 2015

### AJDangles

v = d/t

Simple enough or are you looking for something else?

4. Oct 7, 2015

### haruspex

No, that equation only gives you average velocity over a time period t. Here we are concerned with instantaneous velocity.

5. Oct 7, 2015

### AJDangles

v = delta X / delta T

6. Oct 7, 2015

### AJDangles

would the first answer therefore be, v = (sin (wt +θ)) / ΔT?

edit:

i mean, v = (sin (wt +θ)) / dt?

because we're talking the instan velocity therefore derivative and the slope of the tangent

Last edited: Oct 7, 2015
7. Oct 8, 2015

### haruspex

You're heading in the right direction, but how do you differentiate sin (wt +θ) with respect to t?

8. Oct 8, 2015

### AJDangles

I'm really rusty on my calculus so this could be ugly but here goes...

dv = d (sin(wt + θ)) / dt
dv * dt = d (sinwt + sinθ)
dv * dt = d*sin*wt + d sinθ
dv = (d*sin*wt + dsinθ) / dt
dv = sinw + dsinθ

do i have to integrate or use a trig identity here?

9. Oct 8, 2015

### haruspex

No, v = dx/dt, not dv=dx/dt. The quantities starting with a d are infinitesimals, meaning they will be taken to be arbitrarily close to zero. dx/dt is a ratio of two of these, so need not be tending to zero. So dv=dx/dt makes no sense. The left side is sure to go to zero, the right side is not.
Next, it gains nothing to multiply out by dt. The left hand side, v, is now what is wanted. Your challenge is to do something with dx/dt, i.e. with d (sin(wt + θ)) / dt.
Do you know the derivative of the sine function? Do you remember the chain rule?

10. Oct 8, 2015

### AJDangles

Wooooow it's all coming back. That rush of knowledge. Ok thanks.

I think it got it?

= cos (wt + θ) * d (wt+θ)/dt
= cos (wt + θ) * w
= cosw (wt + θ)
v = cos(w)^2 (t) + cos(w)θ

11. Oct 8, 2015

### AJDangles

and then

a = d (cos (w) (wt + θ)) / dt
a = cos (w) (w + 0)
a = w cos (w)

12. Oct 8, 2015

### haruspex

you were doing well until that last step. Try putting all appropriate parentheses in each equation.

13. Oct 8, 2015

### AJDangles

For the pressure, would the last equation i have to use to derive be
p (t) = (mv)/(t * area)?

14. Oct 8, 2015

### AJDangles

cos (w) (wt + θ) ?

15. Oct 8, 2015

### haruspex

No, that's already wrong. Go back to your post 10 and put all appropriate parentheses in each equation.

16. Oct 8, 2015

### AJDangles

Ok...

v = d/dt (sin (wt + θ))
v = cos (wt + θ) * d/dt (wt + θ) via chain rule.
v = (w * d/dt (t) * d/dt (θ)) * cos (wt + θ)
v = (1w + 0) * cos (wt + θ)
v = w cos (wt + θ) ?

Sorry if my math is rusty; so cosw is different than w cos? Is cosw assumed to be cos (w)?

17. Oct 8, 2015

### haruspex

Yes, much better. Yes, cos is a function. a cos(b) is a completely different beast from cos(ab).